# Sand is pouring from a pipe at the rate of 12cm3/sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone in increasing when the height is 4 cm?

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
7 years ago
Ans:
Let height & radius of cone be at any instant ‘t’ to be ‘h’ & ‘r’.
Volume ‘V’:
$V = \frac{1}{3}\pi r^{2}h$
$r = 6h$
$V = \frac{1}{3}\pi (6h)^{2}h$
$V = 12\pi (h)^{3}$
$\frac{\partial V}{\partial t} = 36\pi h^{2}.\frac{\partial h}{\partial t} = 12 \frac{cm^{3}}{sec}$
$h = 4cm$
$36\pi (4)^{2}.\frac{\partial h}{\partial t} = 12 \frac{cm^{3}}{sec}$
$\frac{\partial h}{\partial t} = \frac{1}{48\pi } \frac{cm}{sec}$
Thanks & Regards
Jitender Singh
IIT Delhi